find all zeros by factoring each function.\n15) ( f(x)=x^{3}-2 x^{2}+x )\n16) ( f(x)=x^{3}+8 )\n17) (…

15)

Explanation:

Step1: Factor out the common factor

Factor out (x) from (f(x)=x^{3}-2x^{2}+x). (f(x)=x(x^{2}-2x + 1))

Step2: Factor the quadratic expression

Use the formula ((a - b)^2=a^{2}-2ab + b^{2}), where (a=x) and (b = 1). So (x^{2}-2x + 1=(x - 1)^{2}). Then (f(x)=x(x - 1)^{2})

Step3: Set (f(x)=0)

Set (x(x - 1)^{2}=0). By the zero - product property, if (ab = 0), then (a=0) or (b = 0). (x=0) or ((x - 1)^{2}=0) Solving ((x - 1)^{2}=0) gives (x = 1) (with multiplicity 2)

Answer:

The zeros of (f(x)) are (x = 0) and (x=1)

16)

Explanation:

Step1: Use the sum of cubes formula

The sum of cubes formula is (a^{3}+b^{3}=(a + b)(a^{2}-ab + b^{2})). For (f(x)=x^{3}+8=x^{3}+2^{3}), where (a=x) and (b = 2) (f(x)=(x + 2)(x^{2}-2x + 4))

Step2: Set (f(x)=0)

Set ((x + 2)(x^{2}-2x + 4)=0) By the zero - product property, (x+2=0) or (x^{2}-2x + 4=0) Solving (x + 2=0) gives (x=-2) For (x^{2}-2x + 4=0), use the quadratic formula (x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}) with (a = 1), (b=-2), (c = 4) (\Delta=b^{2}-4ac=(-2)^{2}-4\times1\times4=4 - 16=-12) (x=\frac{2\pm\sqrt{-12}}{2}=\frac{2\pm2\sqrt{3}i}{2}=1\pm\sqrt{3}i)

Answer:

The zeros of (f(x)) are (x=-2,x = 1+\sqrt{3}i,x = 1-\sqrt{3}i)

17)

Explanation:

Step1: Let (u=x^{2})

Let (u=x^{2}), then (f(x)=x^{4}-x^{2}-30) becomes (y = u^{2}-u - 30)

Step2: Factor the quadratic in (u)

Factor (u^{2}-u - 30=(u - 6)(u + 5)) Substitute back (u=x^{2}), we get (f(x)=(x^{2}-6)(x^{2}+5))

Step3: Set (f(x)=0)

Set ((x^{2}-6)(x^{2}+5)=0) By the zero - product property, (x^{2}-6=0) or (x^{2}+5=0) Solving (x^{2}-6=0) gives (x=\pm\sqrt{6}) Solving (x^{2}+5=0) gives (x=\pm\sqrt{-5}=\pm\sqrt{5}i)

Answer:

The zeros of (f(x)) are (x=\sqrt{6},x=-\sqrt{6},x=\sqrt{5}i,x=-\sqrt{5}i)

18)

Explanation:

Step1: Let (u=x^{2})

Let (u=x^{2}), then (f(x)=x^{4}+x^{2}-12) becomes (y=u^{2}+u - 12)

Step2: Factor the quadratic in (u)

Factor (u^{2}+u - 12=(u + 4)(u - 3)) Substitute back (u=x^{2}), we get (f(x)=(x^{2}+4)(x^{2}-3))

Step3: Set (f(x)=0)

Set ((x^{2}+4)(x^{2}-3)=0) By the zero - product property, (x^{2}+4=0) or (x^{2}-3=0) Solving (x^{2}-3=0) gives (x=\pm\sqrt{3}) Solving (x^{2}+4=0) gives (x=\pm\sqrt{-4}=\pm2i)

Answer:

The zeros of (f(x)) are (x=\sqrt{3},x=-\sqrt{3},x = 2i,x=-2i)

19)

Explanation:

Step1: Use the difference of squares formula

First, (f(x)=x^{6}-64=(x^{3})^{2}-8^{2}) By the difference of squares formula (a^{2}-b^{2}=(a + b)(a - b)), we have (f(x)=(x^{3}+8)(x^{3}-8))

Step2: Use the sum and difference of cubes formulas

For (x^{3}+8=x^{3}+2^{3}=(x + 2)(x^{2}-2x + 4)) For (x^{3}-8=x^{3}-2^{3}=(x - 2)(x^{2}+2x + 4)) So (f(x)=(x + 2)(x - 2)(x^{2}-2x + 4)(x^{2}+2x + 4))

Step3: Set (f(x)=0)

Set ((x + 2)(x - 2)(x^{2}-2x + 4)(x^{2}+2x + 4)=0) (x+2=0) gives (x=-2); (x - 2=0) gives (x = 2) For (x^{2}-2x + 4=0), (x=\frac{2\pm\sqrt{4 - 16}}{2}=1\pm\sqrt{3}i) For (x^{2}+2x + 4=0), (x=\frac{-2\pm\sqrt{4 - 16}}{2}=-1\pm\sqrt{3}i)

Answer:

The zeros of (f(x)) are (x=-2,x = 2,x=1+\sqrt{3}i,x=1-\sqrt{3}i,x=-1+\sqrt{3}i,x=-1-\sqrt{3}i)

20)

Explanation:

Step1: Let (u=x^{3})

Let (u=x^{3}), then (f(x)=x^{6}+2x^{3}+1) becomes (y=u^{2}+2u + 1)

Step2: Factor the quadratic in (u)

Factor (u^{2}+2u + 1=(u + 1)^{2}) Substitute back (u=x^{3}), we get (f(x)=(x^{3}+1)^{2})

Step3: Use the sum of cubes formula

Since (x^{3}+1=(x + 1)(x^{2}-x + 1)), then (f(x)=[(x + 1)(x^{2}-x + 1)]^{2}=(x + 1)^{2}(x^{2}-x + 1)^{2})

Step4: Set (f(x)=0)

Set ((x + 1)^{2}(x^{2}-x + 1)^{2}=0) ((x + 1)^{2}=0) gives (x=-1) (with multiplicity 2) For (x^{2}-x + 1=0), (x=\frac{1\pm\sqrt{1 - 4}}{2}=\frac{1\pm\sqrt{-3}}{2}=\frac{1}{2}\pm\frac{\sqrt{3}}{2}i) (each with multiplicity 2)

Answer:

The zeros of (f(x)) are (x=-1) (multiplicity 2), (x=\frac{1}{2}+\frac{\sqrt{3}}{2}i) (multiplicity 2), (x=\frac{1}{2}-\frac{\sqrt{3}}{2}i) (multiplicity 2)